Calculating Pressure Change in Combustion Cylinder

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In a combustion cylinder with a fixed number of moles of gas, the pressure change when the volume increases from 2.00 L to 6.0 L at constant temperature is calculated using the formula P1V1 = P2V2, resulting in a final pressure (P2) of 3.0 atm and a change in pressure (delta P) of -6.0 atm. For another scenario where the temperature decreases from 9.00 K to 4.00 K with a fixed piston, the pressure change is computed using P1/T1 = P2/T2, yielding P2 of approximately 1.78 atm and a delta P of -2.22 atm. Concerns are raised about the behavior of gases at low temperatures, specifically whether the gas condenses and deviates from ideal gas behavior. If the gas does not behave ideally, the Van Der Waals equation should be applied, but without additional data, the ideal gas law is the default assumption. The calculations appear correct, but the applicability of the ideal gas law at low temperatures is a critical consideration.
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1) A combustion cylinder (containing a fixed number of moles of gas initially at 9.00atm ) with a moveable piston is maintained at a constant temperature changes volume from 2.00 L to 6.0 L. Determine the change in pressure during this action.

P1 = 9.00 atm
V1 = 2.00 L
P2 = ?
V2 = 6.0 L

Use P1V1 = P2V2, where P2 = (P1*V1)/(V2) = (9.00atm*2.00L)/(6.0 L) = 3.0 atm??

delta P (change in P) = P2 - P1 = 3.0 atm - 9.00 atm = -6.0 atm?? (correct significant digits?)

2) A combustion cylinder (containing a fixed number of moles of gas initially with a pressure of 4.00atm ) with a fixed piston changes temperature from 9.00 kelvin to 4.00 kelvin. Determine the change in pressure during this action.

P1 = 4.00 atm
T1 = 9.00 K
P2 = ?
T2 = 4.00 K

Use P1/T1 = P2/T2, where P2 = (P1*T2)/(T1) = (4.00 atm)(4.00 K)/(9. 00K) = 1.7778 atm??

Change in pressure (delta P) = P2 - P1 = 1.78 atm - 4.00 atm = -2.22 atm?

Thank you for any help.
 
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i think it is ok as i don't see any mistakes.
 
The only thing to worry about is whether your gas CONDENSES as it cools to 4 Kelvin. At such low Temperatures, almost nothing acts like an IDEAL gas.
 
Well, exactly what formula/method would you use if it wasn't an ideal gas?
 
Well if he was not given any more information, you must assume it uses the ideal gas law... you would need experimental values otherwise i believe.
 
use the Van Der Waals equation for real gases.
 
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